What to bring and what to study
•One 8.5 X 11 formula sheet, one
side only, no examples. Save the
other side for test 2.
•Put your name on it and turn it in
with the test.
•If you number the formulas.
Suggest you use the same numbers
as the text, you will be free to refer
to them on the test. That is:
“ from equation (1.37): d  20 1 (0.01)2
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 19.9999  20 rad/s "
Mechanical Engineering at Virginia Tech
Material (sections) Covered
• 1.1,1.2, 1.3, 1.4, 1.5, 1.7
• 2.1, 2.2
• Log decrement from 1.6
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Also bring
•
•
•
•
Paper, pencil, calculator
No other resources allowed
Your honor, but no anxiety
Knowledge of all examples worked in
class or presented in the text
• All assigned homework
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Expect 4 to 5 problems
1. An example covered in class
2. A homework problem
3. An example from the book, not
covered in class
4. A problem involving combining parts of
any of the above in “two steps” and/or
5. A derivation
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•25% (or 20%) each
•the last problem(4 and/or 5) intended to
sort out the A’s and B’s
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Free-body diagram and
equations of motion
• Newton’s Law:
mx(t )  kx(t )
mx(t )  kx(t )  0
x (0)  x0 , x (0)  v0
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2nd Order Ordinary Differential
Equation with Constant
Coefficients
Divide by m : x(t )   x (t )  0
2
n
k
n 
natural frequency in rad/s
m
x (t )  A sin( n t   )
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Periodic Motion
amplitude
Displacement
x(0)
Phase
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Maximum
Velocity
2
n
Time usually sec
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Frequency
n is in rad/s is the natural frequency
n rad/s
 cycles n
fn 
 n

Hz
2  rad/cycle
2 s
2
2
T
s is the period
n
We often speak of frequency in Hertz, but we
need rad/s in the arguments of the
trigonometric functions.
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Amplitude & Phase from the ICs
x0  Asin(n 0  )  Asin 
v0  n A cos(n 0  )  n A cos 
Solving yields
A
1
n
n x0 
 x  v ,   tan 

 v0 
2
n
2
0
Amplitude
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2
0
1
Phase
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Other forms of the solution:
x(t)  Asin(n t  )
x(t)  A1 sin  nt  A2 cos n t
x(t)  a1e
j nt
 a2e
 j nt
See window 1.4, page 12 for relationships among these.
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Peak Values
max or peak value of :
displaceme nt : xmax  A
velocity : xmax  n A
accelerati on : xmax   A
2
n
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Spring-mass-damper systems
• From Newton’s law:
mx(t )   f c  f k
 cx (t )  kx(t )
mx(t )  cx (t )  kx(t )  0
x (0)  x0 , x (0)  v0
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Solution: Given m, c, k, x0, v0 find x(t)
Divide equation of motion by m
x(t )  2n x (t )  n2 x (t )  0
where n  k
m
and
c
 =
 damping ratio (dimension less)
2 km
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t
Let x(t)  ae & subsitute into eq. of motion
a e  a2 ne   ae  0
2 t
t
2
n
t
which is now an algebraic equation in  :
1,2  n  n  1
2
from the roots of a quadratic equation
Here the discriminant  1, determines
2
the nature of the roots 
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  1

  0
0    1

Three possibilities:
1)   1  roots are equal & repeated
called critically damped
 = 1  c  ccr  2 km  2mn
x (t )  a1e
n t
 a2te
n t
Using the initial conditions :
a1  x0 , a2  v0  n x0
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2)   1, called overdampin g - two distinct real roots :
1, 2  n  n  2  1
x (t )  e
 n t
where a1 
a2 
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(a1e
n t  2 1
n t  2 1
 a2 e
)
 v0  (    2  1)n x0
2n   1
2
v0  (   2  1)n x0
2n   1
2
Mechanical Engineering at Virginia Tech
3)   1, called underdampe d motion - most common
Two complex roots as conjugate pairs
write roots in complex form as :
1, 2  n  n j 1   2
whe re
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j  1
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Underdamped 0 <  < 1
x (t )  e
 n t
( a1e
jn t 1 2
 a2 e
 jn t 1 2
)
 Ae nt sin( d t   ) = e  nt [ A sin( d t )  B cos(d t )]
d  n 1   2 , damped natural frequency
A
1
( v0  n x0 ) 2  ( x0d ) 2
d
 x0d 

  tan 
 v0  n x0 
1
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Reduces to undamped formulas
for  = 0
Mechanical Engineering at Virginia Tech
Potential and Kinetic Energy
The potential energy of mechanical
systems U is often stored in “springs”
(remember that for a spring F=kx)
x0
Uspring
x=0
x0
k
M
x0
1 2
  F dx   kx dx  kx0
2
0
0
Mass
The potential energy of mechanical
h
systems U is also gravitational: U grav  mgdx  mgh

Spring
0
The kinetic energy of mechanical systems T is due to the
motion of the “mass” in the system
Ttrans
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1
1 2
2
 mx Trot  J
2
2
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Conservation of Energy
For a simply, conservative (i.e. no damper), mass spring
system the energy must be conserved:
T U  constant
or
d
(T U )  0  Equation of motion
dt
At two different times t1 and t2 the increase in potential
energy must be equal to a decrease in kinetic energy (or visaversa).
U1 U 2  T2 T1 and U max  Tmax  An expression for the
natural frequency
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Deriving equation of motion
x=0
x
k
M
Mass
Spring
d
d 1
1 2
2

(T  U )   mx  kx   0
dt
dt  2
2

 x mx  kx   0
Since x cannot be zero for all time
mx  kx  0
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Natural frequency
If the solution is given by Asin(t+) then the maximum potential
and kinetic energies can be used to calculate the natural frequency
of the system
1 2
1
Umax  kA Tmax  m( n A) 2
2
2
Since these two values must be equal
1 2 1
kA  m( n A) 2
2
2
 k  m   n 
2
n
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k
m
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Static Deflection
  distance spring is stretched or
compressed under the force of
gravity by attaching a mass m to it.
mg
  s 
k
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Combining Springs
• Equivalent Spring
A
B
k1
a
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k1
k2
C
k2
b
1
series : kAC  1
1

k1
k2
parallel : kab  k1  k2
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Harmonically Excited Systems
Equations of motion (c =0):
mx(t )  kx(t )  F0 cos(t )
x(t )  n2 x(t )  f 0 cos(t )
where
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f 0  F0 / m, n  k
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m
Linear nonhomogenous ode:
• Solution is sum of homogenous and particular
solution
• The particular solution assumes form of
forcing function (physically the input wins)
x p (t)  X cos( t)
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To be determined
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Driving frequency
Substitute into the equation of motion:
xp
n2 x p

 
 
2
2
  X cos t  n X cos t  f 0 cos t
f0
solving yields : X  2
2
n  
Thus the particular solution has the form:
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f0
x p (t )  2
cos(t )
2
n  
Mechanical Engineering at Virginia Tech
Add particular and
homogeneous solutions to get
general solution:
x(t) 
particular
f0
A1 sin nt  A2 cos nt  2
cos t
2



n
homogeneous
A1 and A2 are constants of integration.
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Apply the initial conditions to evaluate the constants
f0
x (0)  A2  2
 x0
2
n  
x (0)  n A1  v0
Solving for the constants and substituting into x yields

f0 
f0
 cos n t  2
x (t ) 
sin n t   x0  2
cos t
2 
2
n
n   
n  

v0
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2.2 Harmonic excitation of
damped systems
mx(t )  cx (t )  kx(t )  F0 cos t
x(t )  2n x (t )   x(t )  f 0 cos t
2
n
x p (t )  X cos(t   )



now includesa phase shift
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Substitute the values of As and Bs into xp:
 2n 
)
x p (t ) 
cos(t  tan  2
2
n   
(n2   2 )2  (2n )2



f0
1

X
Add homogeneous and particular to get total solution:
x (t )  Ae nt sin( d t   )  X cos(t   )




 
homogeneous or transient solution
particular or
steady state solution
Note: that A and  will not have the same values as in Ch 1,
as t gets large, transient dies out
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Magnitude:
Non dimensional
Form:
Phase:
Frequency ratio:
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f0
X
2
2 2
2
(n   )  (2n )
Xk X
1


F0
f0
(1  r 2 ) 2  (2r) 2
2
n
2r

  tan 
2 
 1  r 
1

r
n
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Magnitude plot
r
0 , 0.01 .. 2
1
X r, 
1
r2
2
2. . r
2
6
X r , 0.1
4
X r , 0.5
X r , 0.7
X r , 0.25
2
0
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0.5
1
r
1.5
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2
Phase plot
 r, 
2. r . 
atan
1
r
2
. 1
r

atan
2. r. 
1
r
2
. r
1
4
 r , 0.1
3
 r , 0.5
 r , 0.7
2
/2
 r , 0.25
z r
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1
0
0.5
1
1.5
r
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2